Differentiation Question 335

Question: If $ x=\frac{3at}{1+t^{3}},y=\frac{3at^{2}}{1+t^{3}}, $ then $ \frac{dy}{dx} $ =

Options:

A) $ \frac{t(2+t^{3})}{1-2t^{3}} $

B) $ \frac{t(2-t^{3})}{1-2t^{3}} $

C) $ \frac{t(2+t^{3})}{1+2t^{3}} $

D) $ \frac{t(2-t^{3})}{1+2t^{3}} $

Show Answer

Answer:

Correct Answer: B

Solution:

$ x=\frac{3at}{1+t^{3}},\ \ \ y=\frac{3at^{2}}{1+t^{3}} $ . Clearly, $ y=tx $ . Differentiate w.r.t. x, we get $ \frac{dy}{dx}=t.\ \ 1+x.\frac{dt}{dx} $ ……..(i) Now, $ \frac{dx}{dt}=3a.\frac{1+t^{3}-t.3t^{2}}{{{(1+t^{3})}^{2}}}=\frac{3a.(1-2t^{3})}{{{(1+t^{3})}^{2}}} $

…..(ii)
$ \therefore \frac{dy}{dx}=t+\frac{3at}{1+t^{3}}.\frac{{{(1+t^{3})}^{2}}}{3a(1-2t^{3})}=\frac{t(2-t^{3})}{1-2t^{3}} $ , {by (ii)}.